What is the extreme f(x)=(x^5)/5-ln(x) ?

The extreme f(x)=(x^5)/5-ln(x) is Minimum(1, 1/5)

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extreme f(x)=(x^5)/5-ln(x)

extreme points

f (x) = \frac{x ^{5}}{5} - ln (x)

Minimum (1, \frac{1}{5})

Solution steps

Find the critical points:

x = 1

Domain of

\frac{x ^{5}}{5} - ln (x) : x > 0

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = 1

into

\frac{x ^{5}}{5} - ln (x) \Rightarrow y = \frac{1}{5}

Minimum (1, \frac{1}{5})

f (x, y) = 8 x^{4} - x^{2} + 3 y^{2}

f (x, y) = 3 x^{2} + y^{3} - 18 x y + 22

f (3.2) = - \frac{40000 t}{( 3 + t ^{2} + 2 x ^{2} ) ^{2}}

\frac{4 e ^{- 2 x}}{2 x + 5}

f (x, y) = - 2 x^{2} - y^{3} + 9 y^{2} + 16 x - 15 y + 5

What is the extreme f(x)=(x^5)/5-ln(x) ?
The extreme f(x)=(x^5)/5-ln(x) is Minimum(1, 1/5)